What is $\lim_{x \to 0} \frac{e^{x - \sin x} - 1}{x - \sin x}$

Question

Evaluate $$\lim_{x \to 0} \frac{e^{x - \sin x} - 1}{x - \sin x}$$

How do I find the limit without L'Hopital's rule?

It is well known that $$\lim_{x \to 0} \frac{e^x - 1}{x} = 1$$ So, if the limit were $$\lim_{(x - \sin x) \to 0} \frac{e^{x - \sin x} - 1}{x - \sin x}$$ the answer would be $1$. But here $x \to 0$, so I have no idea how to proceed.

Any help would be appreciated.

Thanks

Answer 1

Simply note that $$x\to0\implies x-\sin x \to0$$

Answer 2

Set $y=x-\sin x$.

For $x\to 0$ the new variable $y$ tends to $0$. So you can use the well-known limit theorem to conclude that

$$\lim_{x\to0} \frac{e^{x-\sin x}-1}{x-\sin x}=1$$

Answer 3

You already know that $$\lim_{x\to0}\frac{e^x-1}x=1$$

This is true not just for $x$ but for any function of $x$ if that is tending to zero i.e. $$\lim_{f(x)\to0}\frac{e^{f(x)}-1}{f(x)}=1$$

In general, if $f(x)$ is tending to zero when $x$ is tending to some 'a' then we say $$\lim_{x\to a}\frac{e^{f(x)}-1}{f(x)}=1$$

For example, $$\lim_{x\to2}\frac{e^{x-2}-1}{x-2}=1$$

Answer 4

Define $t=x-\sin(x)$, when $x\to0, ~t\to0$

$$\lim_{x\to0} \frac{e^{x-\sin x}-1}{x-\sin x}=\lim_{t\to0}\frac{e^t-1}{t}=\lim_{t\to0}\frac{e^t}{1}=1$$